Nonparametric Methods Questions and Answers

Q: The recorded air quality index (AQI) at a place in a city for 16 consecutive days for the current year is compared with the historical average AQI. The direction of the recorded AQI is noted as '+' if it is above the historical average and as '−' if it is below the historical average. The dataset is as below:+ + − + − − + + + + − − − + + −The total number of runs, \(R\), is being used to test\(H_0\): The direction of the recorded AQI is randomagainst\(H_1\): The direction of the recorded AQI is not randomConsider the following statements.P: The observed value of total number of runs is 8.Q: Based on the above dataset, \(H_0\) is rejected in favor of \(H_1\) at level 0.05.Then which one of the following is true?[You may use: Under \(H_0\), with 7 negative signs and 9 positive signs, \(P(R \leq 6) = 0.108, P(R \leq 7) = 0.231, P(R \leq 8) = 0.427\)]

P is correct and Q is NOT correct.A run is a maximal block of identical signs. Splitting the sequence:(+ +) (−) (+) (− −) (+ + + +) (− − −) (+ +) (−)gives \(R = 8\) runs, with 9 plus and 7 minus signs, so P is correct.The alternative 'not random' is two-sided: too few runs (clustering) or too many (alternation) both count against \(H_{0}\). Under \(H_{0}\) the expected number of runs is \(1 + \frac{2 \cdot 9 \cdot 7}{16} \approx 8.9\), and the observed 8 is close to it. The lower tail \(P(R \leq 8) = 0.427\) is far above the 0.025 a two-sided test at level 0.05 allows each side; even 6 runs would not be significant, since \(P(R \leq 6) = 0.108\). So \(H_{0}\) is not rejected at level 0.05, and Q is not correct.

Q: The following is an arrangement of men (M) and women (W), lined up to purchase tickets for a rock concert :MWMWMMMWMWMMMWWMMMMWWMWMMMWMMMWWWWhat is the expected value of runs for testing the hypothesis that the arrangement is random ?

The expected number of runs is 16.76.Count the letters in the two printed lines: there are 33 people, \(n_{1} = 20\) men and \(n_{2} = 13\) women. Under randomness the number of runs R has mean\[ \begin{aligned} E(R) &= \frac{2n_{1}n_{2}}{n_{1} + n_{2}} + 1 \\ &= \frac{2 \times 20 \times 13}{33} + 1 \\ &= 15.76 + 1 = 16.76 \end{aligned} \](The observed arrangement has 18 runs, close to this expectation, so the runs test would not reject randomness.)None of 14.92, 15.88 or 18.20 is the expectation for any split of 33 people into two groups, since each would need \(n_{1}n_{2}\) to be a non-integer. Dropping the +1 gives 15.76, which is not offered either.

Q: Let \(F_{n}(\cdot)\) be empirical cumulative distribution function, based on a random of size \(n\) from a continuous distribution with cumulative distribution function \(F(\cdot)\). Then :

The asymptotic distribution of \(F_{n}\) is normal.At a fixed \(x\), \(nF_{n}(x)\) counts the observations not exceeding \(x\), which is binomial with parameters \(n\) and \(F(x)\). By the central limit theorem, \(\sqrt{n}(F_{n}(x) - F(x))\) tends to \(N(0, F(x)(1 - F(x)))\).The other statements fail. \(F_{n}(x) \to F(x)\) almost surely (and uniformly, by Glivenko-Cantelli), so \(F_{n}\) is consistent. Its variance is \(F(x)(1 - F(x))/n\), which tends to 0, not to a positive constant. And its mean is \(F(x)\), not 1/2.

Q: \((X_{i}, Y_{i})\) \(i = 1, ......, 9\) is a random sample from bivariate population where all \(X_{i}\)'s are distinct and all \(Y_{i}\)'s are distinct \(i = 1, ......, 9\). If the number of concordance pairs is 9, the Kendall's sample correlation coefficient is :

Kendall's coefficient is \(-1/2\).With 9 observations and no ties there are \(\binom{9}{2} = 36\) pairs, each concordant or discordant. With 9 concordant pairs, 27 are discordant, so\[ \begin{aligned} \tau &= \frac{C - D}{\binom{n}{2}} \\ &= \frac{9 - 27}{36} = -\frac{1}{2} \end{aligned} \]The sign is negative because discordant pairs outnumber concordant ones. +1/2 reverses the sign, and \(\pm 1/4\) come from dividing by 72 or using 9 alone instead of the difference.

Q: Let \(S = S(X_{1}, X_{2}, ......, X_{n})\) be an unbiased estimator of the parametric functional \(\theta(F)\) based on a random sample \(X_{1}, X_{2}, ......, X_{n}\) from F. Suppose U is the U-statistics corresponding to S. Then, which of the following statements is true ?

The true statement is \(V_{F}(U) \le V_{F}(S)\).The U-statistic corresponding to S is its average over all permutations of the sample, which is the conditional expectation of S given the order statistics: \(U = E[S \mid X_{(1)}, \ldots, X_{(n)}]\). The order statistics are sufficient for an i.i.d. sample from an unrestricted F.So U has the same expectation as S, \(\theta(F)\), which rules out A and B, and by the Rao-Blackwell argument (conditional variance formula)\[ V(S) = V(U) + E[V(S \mid \text{order stats})] \ge V(U) \]So the U-statistic never has larger variance, and D is wrong.

Q: X and Y are two continuous random variables independent of each other. The combined sample based on two independent samples of size 5 and 4 on X and Y respectively has the following pattern :\(y_{4} \lt x_{1} \lt x_{5} \lt y_{3} \lt x_{2} \lt y_{1} \lt y_{2} \lt\)\(x_{4} \lt x_{3}\)Hence the value of Mann-Whitney-Wilcoxon test statistic is :

The Mann-Whitney statistic is 8.Count, for each \(y\), the number of \(x\)'s that come before it in the combined ordering \(y_{4} \lt x_{1} \lt x_{5} \lt y_{3} \lt x_{2}\) \(\lt y_{1} \lt y_{2} \lt x_{4} \lt x_{3}\):\(y_{4}\): 0\(y_{3}\): 2 (\(x_{1}, x_{5}\))\(y_{1}\): 3\(y_{2}\): 3The total is \(U = 8\). Counting the other way, the number of \(y\)'s before each \(x\) gives \(1 + 1 + 2 + 4 + 4 = 12\), and \(8 + 12 = 5 \times 4 = 20\) as it must. The test statistic is the smaller count, 8. 12 is the complementary count, and 6 and 10 do not arise from either count.

Q: Let \((X_{i}, Y_{i})\) \(i = 1, \ldots, 5\) be the bivariate pair of observations and \(\mathbf{Z}_{i} = X_{i} - Y_{i}\) \(i = 1, \ldots, 5\). For a given sample the number of \(Z_{i} \gt 0\) is 2. Hence the \(p\)-value to test the null hypothesis that \(H_{0}\) : median (Z) = 0 against \(H_{1}\) : median (Z) \(\lt\) 0 is :

The p-value is 0.5.This is the sign test. Under \(H_{0}\) the number S of positive \(Z_{i}\) is binomial with \(n = 5\), \(p = \frac{1}{2}\). Since \(H_{1}\) says the median is negative, small values of S are evidence against \(H_{0}\), so the p-value is\[ \begin{aligned} P(S \le 2) &= \frac{\binom{5}{0} + \binom{5}{1} + \binom{5}{2}}{32} \\ &= \frac{1 + 5 + 10}{32} = 0.5 \end{aligned} \]The data give no evidence at all against \(H_{0}\). 0.25, 0.10 and 0.05 would require fewer positive differences.

Q: Let \(X_{1}, X_{2}, \ldots, X_{n}\) be iid observations from a continuous distribution F. Define :\[ F_{n}(x) = \begin{cases} 0 & x \lt X_{(1)} \\ k/n & X_{(k)} \le x \lt X_{(k+1)} \\ 1 & x \ge X_{(n)} \end{cases} \]Then for any fixed \(x\), \(\sqrt{n}(F_{n}(x) - F(x)) \xrightarrow{d} Z\), where Z is a normal r.v. with mean 0 and variance :

The variance is \(F(x)(1 - F(x))\).For fixed \(x\), \(nF_{n}(x)\) counts the observations at most \(x\), so it is binomial with parameters \(n\) and \(F(x)\). Hence \(E[F_{n}(x)] = F(x)\) and \(\mathrm{Var}[F_{n}(x)] = F(x)(1 - F(x))/n\).By the central limit theorem, \(\sqrt{n}(F_{n}(x) - F(x)) \to N(0, F(x)(1 - F(x)))\).The other variances, \(F^{2}/2\), \((1 - F)^{2}\) and \(F^{2}\), are not the Bernoulli variance \(p(1 - p)\) with \(p = F(x)\).

Q: Suppose that 12 independent observations on a random variable X are given by 2, 2, 2, 2, 4, 4, 5, 5, 6, 9, 9, 10 and that Y is a random variable with c.d.f. \(F_{12}\), where \(F_{12}\) is the empirical distribution function of the above data. Then :

The true statement is \(P[Y \le 6] = 0.75\).A variable with the empirical distribution function \(F_{12}\) puts mass 1/12 on each of the 12 observations. Nine of them (2, 2, 2, 2, 4, 4, 5, 5, 6) are at most 6, so \(P[Y \le 6] = 9/12 = 0.75\).The others fail. The mean is the sample mean:\[ \begin{aligned} E(Y) &= \frac{8 + 8 + 10 + 6 + 18 + 10}{12} \\ &= \frac{60}{12} = 5 \end{aligned} \]not 4.5. \(P(Y = 7) = 0\) because 7 was not observed, while \(P(Y = 6) = 1/12\). And \(P(Y = 2) = 4/12\) is larger than \(P(Y = 9) = 2/12\), not smaller.

Q: The Mann-Whitney U statistic is defined as :

The Mann-Whitney U statistic is the number of times a Y precedes an X in the combined ordered arrangement of the two samples.Formally \(U = \sum_{i}\sum_{j} I(Y_{j} \lt X_{i})\), the number of \((X, Y)\) pairs in which the Y is smaller. Under the null hypothesis that both samples come from the same continuous distribution, \(E(U) = mn/2\).U is linearly related to the Wilcoxon rank-sum statistic, \(U = W_{X} - m(m + 1)/2\), but the sum of ranks itself (the last option) is the Wilcoxon statistic, not U. The maximum absolute difference between the two empirical distribution functions is the two-sample Kolmogorov-Smirnov statistic, and the minimum difference is not a standard statistic.

The recorded air quality index (AQI) at a place in a city for 16 consecutive days for the current year is compared with the historical average AQI. The direction of the recorded AQI is noted as '+' if it is above the historical average and as '−' if it is below the historical average. The dataset is as below:

+ + − + − − + + + + − − − + + −

The total number of runs, \(R\), is being used to test
\(H_0\): The direction of the recorded AQI is random
against
\(H_1\): The direction of the recorded AQI is not random

Consider the following statements.

P: The observed value of total number of runs is 8.

Q: Based on the above dataset, \(H_0\) is rejected in favor of \(H_1\) at level 0.05.

Then which one of the following is true?

[You may use: Under \(H_0\), with 7 negative signs and 9 positive signs, \(P(R \leq 6) = 0.108, P(R \leq 7) = 0.231, P(R \leq 8) = 0.427\)]

A.P is correct and Q is NOT correct
B.P is NOT correct and Q is correct
C.Both P and Q are correct
D.Neither P nor Q is correct

Exam Year(s)

IIT JAM - 2026,

Answer: A

P is correct and Q is NOT correct.

A run is a maximal block of identical signs. Splitting the sequence:

(+ +) (−) (+) (− −) (+ + + +) (− − −) (+ +) (−)

gives \(R = 8\) runs, with 9 plus and 7 minus signs, so P is correct.

The alternative 'not random' is two-sided: too few runs (clustering) or too many (alternation) both count against \(H_{0}\). Under \(H_{0}\) the expected number of runs is \(1 + \frac{2 \cdot 9 \cdot 7}{16} \approx 8.9\), and the observed 8 is close to it. The lower tail \(P(R \leq 8) = 0.427\) is far above the 0.025 a two-sided test at level 0.05 allows each side; even 6 runs would not be significant, since \(P(R \leq 6) = 0.108\). So \(H_{0}\) is not rejected at level 0.05, and Q is not correct.

Discuss in Forum

The following is an arrangement of men (M) and women (W), lined up to purchase tickets for a rock concert :

MWMWMMMWMWMMMWWMMMMWW
MWMMMWMMMWWW

What is the expected value of runs for testing the hypothesis that the arrangement is random ?

A.14.92
B.15.88
C.16.76
D.18.20

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: C

The expected number of runs is 16.76.

Count the letters in the two printed lines: there are 33 people, \(n_{1} = 20\) men and \(n_{2} = 13\) women. Under randomness the number of runs R has mean

\[ \begin{aligned} E(R) &= \frac{2n_{1}n_{2}}{n_{1} + n_{2}} + 1 \\ &= \frac{2 \times 20 \times 13}{33} + 1 \\ &= 15.76 + 1 = 16.76 \end{aligned} \]

(The observed arrangement has 18 runs, close to this expectation, so the runs test would not reject randomness.)

None of 14.92, 15.88 or 18.20 is the expectation for any split of 33 people into two groups, since each would need \(n_{1}n_{2}\) to be a non-integer. Dropping the +1 gives 15.76, which is not offered either.

Discuss in Forum

Let \(F_{n}(\cdot)\) be empirical cumulative distribution function, based on a random of size \(n\) from a continuous distribution with cumulative distribution function \(F(\cdot)\). Then :

A.\(F_{n}\) is not a consistent estimator for F
B.Variance of \(F_{n}\) converges to a positive constant
C.Asymptotic distribution of \(F_{n}\) is normal
D.Asymptotic mean of \(F_{n}\) is 1/2

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: C

The asymptotic distribution of \(F_{n}\) is normal.

At a fixed \(x\), \(nF_{n}(x)\) counts the observations not exceeding \(x\), which is binomial with parameters \(n\) and \(F(x)\). By the central limit theorem, \(\sqrt{n}(F_{n}(x) - F(x))\) tends to \(N(0, F(x)(1 - F(x)))\).

The other statements fail. \(F_{n}(x) \to F(x)\) almost surely (and uniformly, by Glivenko-Cantelli), so \(F_{n}\) is consistent. Its variance is \(F(x)(1 - F(x))/n\), which tends to 0, not to a positive constant. And its mean is \(F(x)\), not 1/2.

Discuss in Forum

\((X_{i}, Y_{i})\) \(i = 1, ......, 9\) is a random sample from bivariate population where all \(X_{i}\)'s are distinct and all \(Y_{i}\)'s are distinct \(i = 1, ......, 9\). If the number of concordance pairs is 9, the Kendall's sample correlation coefficient is :

A.-1/4
B.-1/2
C.+1/2
D.+1/4

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: B

Kendall's coefficient is \(-1/2\).

With 9 observations and no ties there are \(\binom{9}{2} = 36\) pairs, each concordant or discordant. With 9 concordant pairs, 27 are discordant, so

\[ \begin{aligned} \tau &= \frac{C - D}{\binom{n}{2}} \\ &= \frac{9 - 27}{36} = -\frac{1}{2} \end{aligned} \]

The sign is negative because discordant pairs outnumber concordant ones. +1/2 reverses the sign, and \(\pm 1/4\) come from dividing by 72 or using 9 alone instead of the difference.

Discuss in Forum

Let \(S = S(X_{1}, X_{2}, ......, X_{n})\) be an unbiased estimator of the parametric functional \(\theta(F)\) based on a random sample \(X_{1}, X_{2}, ......, X_{n}\) from F. Suppose U is the U-statistics corresponding to S. Then, which of the following statements is true ?

A.\(E_{F}(U) \gt E_{F}(S)\)
B.\(E_{F}(U) \lt E_{F}(S)\)
C.\(V_{F}(U) \le V_{F}(S)\)
D.\(V_{F}(U) \gt V_{F}(S)\)

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: C

The true statement is \(V_{F}(U) \le V_{F}(S)\).

The U-statistic corresponding to S is its average over all permutations of the sample, which is the conditional expectation of S given the order statistics: \(U = E[S \mid X_{(1)}, \ldots, X_{(n)}]\). The order statistics are sufficient for an i.i.d. sample from an unrestricted F.

So U has the same expectation as S, \(\theta(F)\), which rules out A and B, and by the Rao-Blackwell argument (conditional variance formula)

\[ V(S) = V(U) + E[V(S \mid \text{order stats})] \ge V(U) \]

So the U-statistic never has larger variance, and D is wrong.

Discuss in Forum

X and Y are two continuous random variables independent of each other. The combined sample based on two independent samples of size 5 and 4 on X and Y respectively has the following pattern :

\(y_{4} \lt x_{1} \lt x_{5} \lt y_{3} \lt x_{2} \lt y_{1} \lt y_{2} \lt\)
\(x_{4} \lt x_{3}\)

Hence the value of Mann-Whitney-Wilcoxon test statistic is :

A.6
B.8
C.10
D.12

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: B

The Mann-Whitney statistic is 8.

Count, for each \(y\), the number of \(x\)'s that come before it in the combined ordering \(y_{4} \lt x_{1} \lt x_{5} \lt y_{3} \lt x_{2}\) \(\lt y_{1} \lt y_{2} \lt x_{4} \lt x_{3}\):

\(y_{4}\): 0
\(y_{3}\): 2 (\(x_{1}, x_{5}\))
\(y_{1}\): 3
\(y_{2}\): 3

The total is \(U = 8\). Counting the other way, the number of \(y\)'s before each \(x\) gives \(1 + 1 + 2 + 4 + 4 = 12\), and \(8 + 12 = 5 \times 4 = 20\) as it must. The test statistic is the smaller count, 8. 12 is the complementary count, and 6 and 10 do not arise from either count.

Discuss in Forum

Let \((X_{i}, Y_{i})\) \(i = 1, \ldots, 5\) be the bivariate pair of observations and \(\mathbf{Z}_{i} = X_{i} - Y_{i}\) \(i = 1, \ldots, 5\). For a given sample the number of \(Z_{i} \gt 0\) is 2. Hence the \(p\)-value to test the null hypothesis that \(H_{0}\) : median (Z) = 0 against \(H_{1}\) : median (Z) \(\lt\) 0 is :

A.0.5
B.0.25
C.0.10
D.0.05

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: A

The p-value is 0.5.

This is the sign test. Under \(H_{0}\) the number S of positive \(Z_{i}\) is binomial with \(n = 5\), \(p = \frac{1}{2}\). Since \(H_{1}\) says the median is negative, small values of S are evidence against \(H_{0}\), so the p-value is

\[ \begin{aligned} P(S \le 2) &= \frac{\binom{5}{0} + \binom{5}{1} + \binom{5}{2}}{32} \\ &= \frac{1 + 5 + 10}{32} = 0.5 \end{aligned} \]

The data give no evidence at all against \(H_{0}\). 0.25, 0.10 and 0.05 would require fewer positive differences.

Discuss in Forum

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be iid observations from a continuous distribution F. Define :

\[ F_{n}(x) = \begin{cases} 0 & x \lt X_{(1)} \\ k/n & X_{(k)} \le x \lt X_{(k+1)} \\ 1 & x \ge X_{(n)} \end{cases} \]

Then for any fixed \(x\), \(\sqrt{n}(F_{n}(x) - F(x)) \xrightarrow{d} Z\), where Z is a normal r.v. with mean 0 and variance :

A.\(F^{2}(x)/2\)
B.\((1 - F(x))^{2}\)
C.\(F^{2}(x)\)
D.\(F(x)(1 - F(x))\)

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: D

The variance is \(F(x)(1 - F(x))\).

For fixed \(x\), \(nF_{n}(x)\) counts the observations at most \(x\), so it is binomial with parameters \(n\) and \(F(x)\). Hence \(E[F_{n}(x)] = F(x)\) and \(\mathrm{Var}[F_{n}(x)] = F(x)(1 - F(x))/n\).

By the central limit theorem, \(\sqrt{n}(F_{n}(x) - F(x)) \to N(0, F(x)(1 - F(x)))\).

The other variances, \(F^{2}/2\), \((1 - F)^{2}\) and \(F^{2}\), are not the Bernoulli variance \(p(1 - p)\) with \(p = F(x)\).

Discuss in Forum

Suppose that 12 independent observations on a random variable X are given by 2, 2, 2, 2, 4, 4, 5, 5, 6, 9, 9, 10 and that Y is a random variable with c.d.f. \(F_{12}\), where \(F_{12}\) is the empirical distribution function of the above data. Then :

A.E(Y) = 4.5
B.\(P[Y \le 6] = 0.75\)
C.P(Y = 7) = P(Y = 6)
D.\(P(Y = 2) \le P(Y = 9)\)

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: B

The true statement is \(P[Y \le 6] = 0.75\).

A variable with the empirical distribution function \(F_{12}\) puts mass 1/12 on each of the 12 observations. Nine of them (2, 2, 2, 2, 4, 4, 5, 5, 6) are at most 6, so \(P[Y \le 6] = 9/12 = 0.75\).

The others fail. The mean is the sample mean:

\[ \begin{aligned} E(Y) &= \frac{8 + 8 + 10 + 6 + 18 + 10}{12} \\ &= \frac{60}{12} = 5 \end{aligned} \]

not 4.5. \(P(Y = 7) = 0\) because 7 was not observed, while \(P(Y = 6) = 1/12\). And \(P(Y = 2) = 4/12\) is larger than \(P(Y = 9) = 2/12\), not smaller.

Discuss in Forum

10.

The Mann-Whitney U statistic is defined as :

A.The minimum absolute difference between the two empirical distribution functions
B.The maximum absolute difference between the two empirical distribution functions
C.The number of times a Y precedes X in the combined ordered arrangement of two independent random samples on X's and Y's
D.The sum of ranks of Y's in the combined ordered arrangement of two independent random samples on X's and Y's

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: C

The Mann-Whitney U statistic is the number of times a Y precedes an X in the combined ordered arrangement of the two samples.

Formally \(U = \sum_{i}\sum_{j} I(Y_{j} \lt X_{i})\), the number of \((X, Y)\) pairs in which the Y is smaller. Under the null hypothesis that both samples come from the same continuous distribution, \(E(U) = mn/2\).

U is linearly related to the Wilcoxon rank-sum statistic, \(U = W_{X} - m(m + 1)/2\), but the sum of ranks itself (the last option) is the Wilcoxon statistic, not U. The maximum absolute difference between the two empirical distribution functions is the two-sample Kolmogorov-Smirnov statistic, and the minimum difference is not a standard statistic.

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